FURTHER SOLUTIONS OF THE FALKNER-SKAN EQUATION

FURTHER SOLUTIONS OF THE FALKNER-SKAN EQUATION LAZHAR BOUGOFFA a, RUBAYYI T. ALQAHTANI b Department of Mathematics and Statistics, College of Science, Al-Imam Mohammad Ibn Saud Islamic University (IMSIU), Riyadh, Saudi Arabia E-mail a : lbbougoffa@imamu.edu.sa E-mail b : rtalqahtani@imamu.edu.sa Received April 8, 17 Abstract. In this paper, we propose a reliable treatment for the Falkner-Skan equation, which can be described as the non-dimensional velocity distribution in the laminar boundary layer over a flat plate. We propose an algorithm of two steps that will introduce an exact solution to the equation, followed by a correction to that solution. Three different special cases: Hiemenz flow (β = 1), Homann flow (β = 1 ), and Blasius problem (β = ) have been considered. When the pressure gradient parameter β takes sufficiently large values, we use a transformation of variable that reduces the Falkner- Skan equation into an equivalent boundary value problem in a finite domain, and we solve this problem by a different technique that allows us to find f (). Also, various exact solutions can be obtained in a straightforward manner by using a direct method when β = 1. The new technique, as presented in this paper, has been shown to be very efficient for solving the Falkner-Skan equation. Key words: Falkner-Skan equation; Hiemenz flow; Homann flow; Blasius problem; approximate solution; exact solution. 1. INTRODUCTION The Falkner-Skan equation describes the class of the so-called similar laminar flows in boundary layer on a permeable wall, and at varying main - stream velocity [1, ]. This well-known nonlinear third-order differential equation is given by subject to the boundary conditions and f + ff + β ( 1 (f ) ) =, < η <, (1) f() = γ, f () = () f ( ) = 1, (3) where f is a stream function, η is a distance from the wall, which is called similarity variable. f defines the velocity component in η-direction, f defines the shear stress in the boundary layer. The mass-transfer parameter γ in the boundary condition sets the measure for the Romanian Journal of Physics 63, 1 (18) v..1*18..16#cfdcf3f

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani mass flow rate through the wall boundary in either direction. Positive values determine flows with suction, negative with blowing through the wall boundary. The zero value corresponds to flow along impermeable wall with zero mass transfer. The numerical parameter β (positive or negative) in the Falkner-Skan equation sets a degree of acceleration or deceleration of main stream [3]. The Falkner-Skan equation (1), together with ()-(3), constitute the theoretical basis of convective heat and mass transfer. The existence and uniqueness question for the problem (1)-(3) was given by [4] and [5]. Weyl [6] proved that for each value of the parameter β there exists a physical solution with a positive monotonically decreasing, in [, 1), second derivative that approaches zero as the independent variables goes to infinity. In the case β > 1, Craven and Pelietier [7] computed solutions for which df dη < for some values of η. Coppel [8] presented an elegant proof of the existence and uniqueness for β > and showed that the wall shear stress f () is an increasing function of β. In the works [9, 1], the extension for β < was discussed and properties of solutions were investigated further. New theoretical results for the solutions of the Falkner-Skan equation were given in Refs. [1] and [11]. Different methods have been proposed to solve this problem: finite difference [9, 1 14], shooting [15 ], finite element [1], group invariance theory [, 3], Chebyshev spectral method [4, 5], differential transformation method [6], non-iterative transformation method (ITM) [7], homotopy analysis method [8], and Adomian decomposition method (ADM) [9]. Other direct methods and numerical techniques to study nonlinear differential equations of physical relevance have been recently reported [3 3]. To our knowledge, a general analytic solution of Eq. (1) satisfying the boundary conditions ()-(3) does not exist. The purpose of this work is to present an algorithm of two steps that will introduce approximate solutions to the Falkner-Skan equation followed by a correction to that solution. Furthermore, we will consider three special cases: Hiemenz flow (β = 1), Homann flow (β = 1 ) and Blasius problem (β = ). We use a transformation of variables that reduces (1)-(3) into an equivalent boundary value problem (BVP) in a finite domain, Finally, we solve this problem by a different technique that allows us to find f (), when the pressure gradient parameter β takes sufficiently large values. Also, various exact solutions can be obtained in a straightforward manner by using a direct method when β = 1. The results lead to very accurate approximate solutions. Listed below are some special cases when Eqs. (1)-(3) are solvable.

3 Further solutions of the Falkner-Skan equation Article no. 1. CASE 1: LARGE VALUES OF β The Falkner-Skan can be written in an equivalent form utilizing m as the Falkner-Skan pressure parameter instead of β : f + ff + m ( 1 (f ) ) =, (4) m + 1 where m. The authors have considered only the case of β. For large values of m, Eq. (1) becomes f + ff + ( 1 (f ) ) =. (5) This case has been studied in Ref. [5]. One of the main problem appears when β. In this section, we will present a direct approach to solve this problem when the pressure gradient parameter β takes sufficiently large values, i.e. β. Let us consider the following transformation [33]: Differentiating Eq. (1) with respect to η, we get and from Eq. (6), we have and z(x) = f (η) and x = f (η). (6) f + ff + (1 β)f f = (7) z(x) = df dη = dx dη, (8) z (x) = dz dx = f f = f β 1 f f (9) z (x) = d z dx = f f. (1) f f 3 Substituting these equations into Eq. (7), we obtain z (x) β(1 x ) z (x) 1 z + (1 β)x =. (11) (x) z(x) Since f ( ) = 1 and f() =, dealing, for example, with a simple case γ =, we have z(1) = f ( ) = and z () = β σ and z() = σ, where f () = σ. Hence, the initial-boundary value problem for the Falkner-Skan equation can be transformed into the new problem { z (x) β(1 x ) z (x) 1 + (1 β)x z (x) z(x) =, < x < 1, z() = σ,z () = β σ,z(1) =. (1)

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 4 (1 β)x Multiplying both sides of the first-equation of (1) by ξ(x) = e β(1 x dx ) and taking into account that ξ (x) = (1 β)x ξ(x), we obtain β(1 x ) ( ) ξ(x)z (x) + β(1 x 1 ) ξ(x) =. (13) z(x) We have ξ(x) = e (1 β) x β 1 x dx = (1 x ) 1 1 β. (14) Thus ( ) z (x) + β(1 x ) 1 1 β ξ(x) =. (15) z(x) The function (1 x ) 1 β can be approximated by 1 for sufficiently large values of β, and then ξ(x) 1 x. More precisely, we obtain the approximated equation ( ) z 1 (x) + β ξ(x) =. (16) z(x) Integrating Eq. (16) from to x and taking into account that z () = β σ and z() = σ, we get z 1 (x) + βξ(x) =, (17) z(x) which can be written as follows (z (x)) = βξ(x). (18) Integrating Eq. (18) from to x and taking into account that z() = σ, we get Consequently, z (x) = β x ξ(x)dx + σ. (19) z (x) = β(x x3 3 ) + σ. () Thus, it is required to find σ by the given condition z(1) =. Then we have σ = 4 β. (1) 3 Thus It follows that β f () = σ = ± 3. () z(x) = ± β(x x3 3 ) + 4 β. (3) 3

5 Further solutions of the Falkner-Skan equation Article no. 1 3. CASE : HOMANN FLOW (β = 1 ) For β = 1, Eq. (1) becomes { z (x) 1 (1 x ) z (x) =, z (x) z() = σ,z () = β σ,z(1) =. (4) We start first by noting that the assumption z(x) = k, (5) where k is a parameter to be determined, satisfies the first equation of (4). Using the boundary condition z() = σ, we obtain k = σ. However, this solution does not satisfy both boundary conditions z () = β σ and z(1) =. In view of this, we will employ the given conditions together with (5) to achieve the second goal of our approach by making a correction to (5). We do this by writing the first equation of (4) in an equivalent form as z (x) = 1 (1 x ) z (x) z (x). (6) Inserting the obtained solution (5) in the right hand side (RHS) of Eq. (6), we obtain z(x) = ax + b, (7) where a and b are two constants of integration. Taking into account the boundary conditions z() = σ, z () = β σ = 1 σ, z(1) =, we immediately find that a = 1 σ, b = σ (8) with σ = 1 =.771. The value of the skin friction for Homann stagnation was determined to be σ =.9768 [15]. Returning to the original transformation, we get so that and f (η) = af (η) + b, (9) f (η) = ce aη b a f(η) = c a eaη b η + d, (31) a where c and d are two constants of integration. Using the boundary conditions ()- (3), we obtain c = σ and d = 4σ 3. Consequently, the approximate solution of the Homann flow problem is given as (3) f p (η) = 4σ 3 e 1 σ η σ η 4σ 3, σ =.771. (3)

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 6 3.1. THE ERROR REMAINDER An error analysis ER(η) for f p (η) is considered in Table 1, where the residual ER(η) or error remainder is calculated for different values of η by substituting f p (η) for f(η) in Eq. (1) when β = 1 : ER 1 (η) = 1 ( e 1 σ η + σe 1 σ η 4σ 3 e 1 σ η + σ η 4σ 3) + 1 [ ( 1 4σ 4 1 e 1 η) ] σ. (33) This demonstrates that the approximate solution is a good approximation of the actual solution. The complete error analysis is given in Tables 1-4. 4. CASE 3: HIEMENZ STAGNATION FLOW (β = 1) We note that the solution of the form f(η) = βη + λ, (34) where λ is a parameter to be determined, satisfies Eq. (1). Then we have Inserting this solution into Eq. (1) to get f (η) = β,f (η) = f (η) =. (35) β(1 β ) =. (36) Thus β = 1, 1,. For the Hiemenz flow problem (β = 1), we have f(η) = η + λ. (37) Inserting now the boundary condition f() = γ we get λ = γ. We write Eq. (1) in an equivalent form as f + ff = ( 1 (f ) ). (38) Inserting the obtained solution Eq. (37) in the RHS of Eq. (38), we obtain f f = f (39) Integrating Eq. (39) twice, we obtain ( ) π f (η) = c e γ η + γ erf + d, (4) where erf is the error function given by erf(x) = x e t dt. (41) π

7 Further solutions of the Falkner-Skan equation Article no. 1 Consequently, π f p (η) = c e γ [ (η + γ)erf ( ) ] η + γ (η+γ) + π e + dη + e, (4) where c,d, and e are constants of integration. Using the boundary conditions f () = and f ( ) = 1, we obtain ( ) π c e γ γ erf + d = (43) and π c e γ + d = 1. (44) Solving this system for c and d, we obtain and 1 c = π e γ ( ) (45) π e γ erf γ d = π e γ erf ( γ ( π e γ erf γ The constant e can be determined from the condition f() = γ. Thus π e = γ c e γ [ ) ) π e γ ( ) γ γerf +. (46) γ π e ]. (47) Hence the approximate solution of the Hiemenz flow problem (β = 1) is given by Eq. (4). Figures 1 and show the approximate solution. 4.1. THE ERROR REMAINDER An error analysis ER(η) for ( ) η f p (η) = ηerf + η π e π when γ = is considered in Table 3, where the residual ER (η) or error remainder is calculated for different values of 1 η <. [ ER (η) = η π e π η + ηerf( η ] ( ) + η π e +1 erf( η )). π (49) (48)

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 8.5 f p (η) 1.5 1.5.5 1 1.5.5 η Fig. 1 The approximate solution of the Hiemenz flow problem f p (η) when γ =. Table 1 The error remainder ER 4 (η) for the value of γ =. η ER 4 (η) -.1951348 1 -.717891.4318148 5.133994 1.39454.11118685 5.5341368 1 8 1.81156498 1 17.375913644 1 36 5.65174347 1 93 1.691376971 1 189.1799367855 1 38 5.13113185 1 955

9 Further solutions of the Falkner-Skan equation Article no. 1 (a) β = 5 Approximation solution Numerical solution f p (η) 15 1 5 (b) β =.5 5 1 15 5 5 η Approximation solution Numerical solution f p (η) 15 1 5 (c) β = 1 5 5 1 15 5 η Approximation solution Numerical solution f p (η) 15 1 5 5 1 15 5 η Fig. The approximate solution compared to numerical solution, γ =.

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 1 (a) η 15 α=. α=.5 α=1. α=1.5 α=. f p (η) 1 5 5 1 15 (b) η 5 5 4 α=. α=.5 α=1. α=1.5 α=. η 3 f p (η) 1 1 3 4 5 η Fig. 3 The approximate solution of the Blasius equation f p (η) when γ = and α =,.5, 1, 1.5,.

11 Further solutions of the Falkner-Skan equation Article no. 1 (a) η 3 3 5 α=. α=-.5 α=-1. α=-1.5 α=-. f p (η) 15 1 5 (b) η 5 4 3 1 5 1 15 5 3 α=. α=-.5 α=-1. α=-1.5 α=-. η f p (η) -1 - -3-4 1 3 4 5 η Fig. 4 The approximate solution of the Blasius equation f p (η) when γ = and α =,.5, 1, 1.5,.

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 1 (a) η 3 3 5 γ=. γ=.5 γ=1. γ=1.5 γ=. f p (η) 15 1 5 5 1 15 5 3 (b) η 5 5 4 γ=. γ=.5 γ=1. γ=1.5 γ=. η 3 f p (η) 1 1 3 4 5 η Fig. 5 The approximate solution of the Blasius equation f p (η) when α = and γ =,.5, 1, 1.5,.

13 Further solutions of the Falkner-Skan equation Article no. 1 Table The error remainder ER 3 (η) for different value of γ. η ER 3 (η), γ=.5 ER 3 (η), γ=1 ER 3 (η), γ=1.5 ER 3 (η), γ= 1.367879441.13533583.497876837.1831563889.5314744.1536591.565855.18651115 5.335779415.699718793.164338.1816178 1 3 1.668699.486149.7531978 1 5.1855386 1 7.866766.391619188 1 7.1777948364 1 11.87187384 1 16 5.6859491 1 9.94587455 1.13153111 1 3.188378 1 41 1.1994635 1 19.36887516 1 41.71334513 1 63.1375756 1 84.1994635 1 19.75395489 1 84.14491844 1 17.3811187498 1 171 5.133195918 1 15.355516367 1 14.948947965 1 33.539349 1 431 1.7117451831 1 14.5788939 1 431.36178994 1 648.573959337 1 865.114684184 1 43.5154951 1 865.61437199 1 199.13743567 1 1733 5.91765767 1 18.16845115 1 167.39916 1 353.567683841 1 4337 Table 3 The error remainder ER (η) for the value of γ =. η ER (η) 1 1.3657348.48564534 5.177934859 1 5 1.18738194 1 1.36964684 1 86 5.1153795 1 541 1.54985463 1 17.477397794 1 8684 5.13835871 1 5484 1.9346463 1 17145 5 5. CASE 4: BLASIUS FLOW (β = ) 5.1. γ For β =, we have f(η) = γ. (5) Proceeding as before, inserting the obtained solution Eq. (5) in the RHS of Eq. (39), we obtain f(η) = c γ e γη + dη + e, γ. (51)

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 14 Table 4 The error remainder ER 1 (η) for the value of γ =. η ER 1 (η) 1.367838.51817878 5.8899848 1 1 1.55999659 1.91899948 1 4 5.1917981679 1 4 1.1917981664 1 4.1917981664 1 4 5.1917981664 1 4 1.1917981664 1 4.1917981664 1 4 5.1917981664 1 4 Using the boundary conditions ()-(3), we obtain c = γ, d = 1, e = γ 1 γ. (5) Consequently, the approximate solution of Eqs. (1)-(3) when β = is given as f p (η) = 1 γ e γη + η + γ 1, γ. (53) γ If f () = α, where α is a constant, then c = γ(1 α). The initial condition f () = α indicates the slip condition at the wall. The case where α = represents no-slip. Proceeding as before, we obtain 5.. γ = f(η) = c k e kη + dη + e. (54) Using the boundary conditions ()-(3), we obtain c = k, d = 1, e = 1 k. (55) Consequently, the approximate solution of Eqs. (1)-(3) when β = is given as f p (η) = 1 k e kη + η 1 k. (56)

15 Further solutions of the Falkner-Skan equation Article no. 1 To find the value of k, we use the transformation [33]: { z(η)z (η) + η =, z () =,z(1) = (57) with z() = σ. Proceeding as in [34], we define the solution z(η) by an infinite series in the form Thus z(η) = C n η n. (58) n= z (η) = (n + 1)C n+1 η n, (59) n= z (η) = η n n= k= n C k C n k (6) and z (η) = η n n= k= n (k + 1)(n + 1 k)c k+1 C n+1 k. (61) The substitution into Eq. (57) yields η n n= k= n n C k C n k = σ 1 3 η3 η n (k + 1)(n 1 k)c k+1 C n 1 k. (6) n(n 1) n= k=

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 16 We equate the coefficients of the like powers of η on the left side and on the right side to arrive at recurrence relations for the coefficients. Thus C = σ, 1 k= C kc 1 k =, k= C kc k = C 1, 3 k= C kc 3 k = 1 3 + 1 1 3 k= (k + 1)( k)c k+1c k, 4 k= C kc 4 k = + 1 3 k= (k + 1)(3 k)c k+1c 3 k, (63) 5 k= C kc 5 k = + 1 3 6 k= (k + 1)(4 k)c k+1c 4 k, 6 k= C kc 6 k = + 1 4 6 k= (k + 1)(5 k)c k+1c 5 k,... The coefficients C,C 1,... are C = σ, C 1 = C =, C 3 = 1 6σ, C 4 = C 5 =, C 6 = 1 18σ 3,... (64) We can now compute the truncated decomposition series z(η) = z (η) + z 1 (η) + z (η) + z 3 (η) + z 4 (η) + z 5 (η) + z 6 (η). Consequently, the solution is given by z(η) = σ η3 6σ η6 18σ 3. (65) Thus, it is required to find σ by the given condition z(1) = ; then we get σ 1 6σ 1 =. (66) 18σ3

17 Further solutions of the Falkner-Skan equation Article no. 1 Then, we get σ =.4417, that is f p () =.44174. The numerical value of f () by the Runge-Kutta method is given as f () =.4696. Finally, substituting this value into Eq. (56), we obtain k =.44174. Consequently, the approximate solution of the initial-boundary value problem for the Blasius equation when γ =, which is characterized by k =.44174, is given by Eq. (56). 5.3. THE ERROR REMAINDER The error remainder ER(η) is calculated for different values of η : 5.3.1. Case 1: γ ER 3 (η) = e γη + e γη (η 1). (67) 5.3.. Case : γ = ER 4 (η) = e kη + e kη (kη 1 k ), k =.44174. (68) where the residuals ERi(x), i = 3,4 or errors remainders are calculated for small and large values of η, which show that the present solutions are highly accurate. The figures 3-5 have been drawn to show the approximate solutions of the Blasius equation. 6. EXACT SOLUTIONS OF THE FALKNER-SKAN EQUATION WHEN β = 1 The solutions for the value of β < represent decelerating flows. We integrate Eq. (1) from to η, using the fact that The substitution of Eq. (69) into Eq. (1) leads to ff = (ff ) f. (69) f + (ff ) = 1. (7) Integrating Eq. (7) from to η and taking into account that f() = γ and f () =, we obtain f + ff = η + f (). (71) Consequently Eq. (71) can also be integrated analytically, and we obtain the nonlinear Riccati equation f + 1 f = 1 η + f ()η + γ. (7)

Article no. 1 Lazhar Bougoffa, Rubayyi T. Alqahtani 18 The first result in the analysis of Riccati equation (7) is: if one particular solution to Eq. (7) can be chosen as follows f 1 = η + f (), then f () must satisfy that is and the general solution is obtained as f () + = γ, (73) f () = γ, γ (74) f(η) = f 1 (η) + 1 w(η), (75) where w satisfies the corresponding first - order linear equation w f 1 w = 1, (76) and we can obtain its closed form solution ( ) π e f () erf η+f () + C w(η) =, (77) e η f ()η where erf is the error function and C is a constant of integration. A set of solutions to the Riccati equation is then given by f(η) = η + f () + e η f ()η ( ), (78) π e f () erf η+f () + C and therefore f satisfies the boundary condition (3), that is f ( ) = 1. Inserting now the boundary condition f() = γ into Eq. (78) to get f () + ( ) = γ. (79) π e f () erf f () + C Consequently, this leads to 1 C = γ f () 1 ( π e f () f () erf ). (8) Thus we have proved. Lemma 1 The solution of the Falkner-Skan equation when β = 1 subject to the boundary conditions ()-(3) can be exactly obtained by Eq. (8), where C is determined by Eq. (79) and γ.

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